A question about eleastic potential energy

[darksoulzero]

Homework Statement

A stretched elastic band of mass 0.55 g is released so that its initial velocity is horizontal, and its initial position is 95 cm above the floor. What was the elastic potential energy stored in the stretched band if, when it lands, it has a horizontal displacement of 3.7 m from the initial position under negligible air resistance?

Homework Equations

d_y=v_1yt+.5a_yt²
E_e=.5kx²
E_k=.5mv²

The Attempt at a Solution

Since I knew that it would take the same amount of time for the elastic band to fall to cover the range I used the equation d_y=V_1yt+ 0.5a_y(t)² and rearranged it in terms of t. I got 0.4403 s. I then found v_x by dividing 3.7m by 0.4403s. Since air resistance is negligible I assumed that all of the elastic potential energy, E_e, would equal the kinetic energy. I then used the equation E_k=.5mv² and determined the E_e to be 19.41818421 J.

 

[rock.freak667]

I didn't recheck your numbers, but how you went about solving it is correct. (I assumed you solved for t when d_y = 0.95m)

 

[PhanthomJay]

Your equations are very good, but your math is off by several decimal points. Then be sure to round it off to 2 significant figures.

 

[ppkjref]

dy=V1yt+ 0.5ay(t)2 and rearranged it in terms of t. I got 0.4403 s.

how do you come to 0.4403 s. if there are two unknown variables (v1 and t)?

 

[ppkjref]

nevermind I'm an idiot v1 = 0